Question 3.8: Calculate the temperature of the rectangular body of Example......

The temperatures of the rectangle at the surface points and time of interest may be derived by using the fdX22By_- t10Y22B00T0_pua.m function. The results are given in Table 3.8 where, for the numerical values of the “building block” temperature, \lambda_{j}(0,\tilde{y}_{P},\tilde{t}_{P}), the first ten decimal-places have been considered (A = 10). In addition, the numerical values of temperature in °C given in the table are obtained by using the following equation that relates the dimensional and dimensionless temperatures:

T(x_{P},y_{P},t_{P})=T_{i n}+\underbrace{\left(\tilde{x}_{P},\tilde{y}_{P},\tilde{t}_{P},\tilde{W},P,A,N\right)}_{={\mathrm{fd}}X22By_{-}t10Y22B00T0_{-}pua} \times \sigma _q(\frac{L}{K} )L^p (3.82)

where \tilde{x}_{P}=0,\,\tilde{t}_{P}=0.05,\,\tilde{W}=4,\,\mathrm{and}\ p=1.

Discussion:
It is interesting to observe that the numerical value of temperature does not change at the middle point of the heated surface (10 cm) when reducing the space step (as expected) according to (i) linear-in-space variation of the surface heat flux, and (ii) middle point of the boundary. This is an application of “numerical” intrinsic verification of the fdX22By_t10Y22B00T0_pua.m function (Cole et al. 2014; D’Alessandro and de Monte 2018). Another observation concerns the same numerical value of temperature at 1, 10, and 19 cm when dealing with only one space step, that is, N = 1. In such a case, in fact, the heating is uniform and the 2D problem reduces to a 1D one, that is, X22B10T1 (in other words, the temperature is independent on y, as expected). This is another application of “numerical” intrinsic verification of the MATLAB function.